Aberrations — chromatic, spherical (concepts): Exercises

Level 1 — Recognition

L1.1 — Which colour focuses closest?

A simple converging glass lens is illuminated with white light. Blue, yellow, and red light each form a focus on the axis. Rank the three focal distances from lens, smallest first.

Recall Solution

WHAT we use: $f \propto \frac{1}{n - 1}$ from the lensmaker's equation. WHY: glass is dispersive, so $n_{F} > n_{D} > n_{C}$ (blue index biggest, red smallest). A bigger $n$ means a bigger $(n - 1)$ , hence a smaller $f$ . Answer: blue closest, then yellow, then red farthest: $f_{F} < f_{D} < f_{C} .$

L1.2 — Name that defect

For each statement say whether it describes chromatic (C) or spherical (S) aberration.

Appears even for a single wavelength.
Caused by $n$ changing with $λ$ .
Cured by stopping down the aperture.
Cured by an achromatic doublet.
Worst for rays far from the axis.

Recall Solution

S — spherical is monochromatic (parent §2).
C — colour dependence is the definition of chromatic.
S — stopping down removes marginal rays, the SA offenders (not the colour split).
C — the doublet balances two dispersive powers.
S — SA grows with ray height $h$ ; it is a marginal-ray effect.

Level 2 — Application

L2.1 — Longitudinal chromatic spread

A thin crown lens has $f = 200 mm$ , with $n_{C} = 1.514$ , $n_{D} = 1.517$ , $n_{F} = 1.522$ . Find (a) the dispersive power $ω$ and (b) the axial separation $f_{C} - f_{F}$ between the red and blue foci.

Recall Solution

(a) WHAT: apply $ω = \frac{n _{F} - n _{C}}{n _{D} - 1}$ . WHY: $ω$ is the fractional focal spread per unit of $f$ . $ω = \frac{1.522 - 1.514}{1.517 - 1} = \frac{0.008}{0.517} = 0.01547.$ (b) WHAT: use $\frac{f _{C} - f _{F}}{f} = ω$ (parent's boxed longitudinal result). WHY: it converts the fractional spread into a real length. $f_{C} - f_{F} = ω f = 0.01547 \times 200 = 3.09 mm .$ Blue's focus sits ~3.1 mm nearer the lens than red's.

L2.2 — Focal shift from the boxed formula

Using $\frac{df}{f} = - \frac{d n}{n - 1}$ , if going red→blue raises the index by $d n = + 0.006$ for a lens with $n - 1 = 0.52$ and $f = 150 mm$ , find $df$ and say which way $f$ moves.

Recall Solution

WHAT: plug into the boxed differential. WHY: it directly relates a small index change to a small focal change. $df = - f \frac{d n}{n - 1} = - 150 \times \frac{0.006}{0.52} = - 1.73 mm .$ The minus sign means $f$ shrinks — the bluer light focuses closer, exactly as expected.

Level 3 — Analysis

L3.1 — Design an achromatic doublet

You want a doublet of total focal length $F = + 500 mm$ ( $P = + 2.0 D$ , taking $F$ in metres). Crown glass has $ω_{1} = 0.021$ , flint has $ω_{2} = 0.038$ . Find the individual powers $P_{1}$ (crown) and $P_{2}$ (flint), and their focal lengths.

Recall Solution

WHAT we set up: two thin lenses in contact obey $P = P_{1} + P_{2}$ , and the achromatic condition is $ω_{1} P_{1} + ω_{2} P_{2} = 0$ (parent's boxed result). See Achromatic Doublet. WHY two equations: we have two unknowns $P_{1}, P_{2}$ ; one equation fixes the power, the other fixes the colour correction. $P_{1} + P_{2} = 2.0, ω_{1} P_{1} + ω_{2} P_{2} = 0.$ From the second, $P_{2} = - \frac{ω _{1}}{ω _{2}} P_{1} = - \frac{0.021}{0.038} P_{1} = - 0.5526 P_{1}$ . Substitute into the first: $P_{1} (1 - 0.5526) = 2.0 \Rightarrow P_{1} = \frac{2.0}{0.4474} = + 4.470 D,$ $P_{2} = 2.0 - 4.470 = - 2.470 D .$ Focal lengths: $f_{1} = 1/ P_{1} = + 0.2237 m = + 223.7 mm$ (converging crown), $f_{2} = 1/ P_{2} = - 0.4049 m = - 404.9 mm$ (diverging flint). Sanity: $P_{1} > 0$ , $P_{2} < 0$ — opposite signs, as forced by $ω > 0$ . ✓

L3.2 — Why marginal rays cross early

Using the expansion $sin θ = θ - \frac{θ ^{3}}{6} + \dots$ , explain in one line each: (a) why the leading correction is third-order, (b) why it makes marginal rays bend more, (c) how the blur scales with aperture radius $h$ .

Figure — Aberrations — chromatic, spherical (concepts)

Recall Solution

(a) The next term after $θ$ is $- θ^{3} /6$ — the first neglected term in the paraxial (small-angle) formula, hence "third-order" or Seidel. There is no $θ^{2}$ term because $sin$ is odd. (b) The true $sin θ$ is smaller than $θ$ , so Snell's law delivers slightly more bending toward the axis than the linear formula predicts. Marginal rays (large $θ$ ) feel this most and cross the axis sooner → focus nearer. (c) Since $θ \propto h$ (ray height) in the Paraxial Approximation regime, the $θ^{3}$ term scales as $h^{3}$ . Look at the figure: the outer (coral) rays cross at the near point, the inner (mint) rays cross at the paraxial focus.

Level 4 — Synthesis

L4.1 — Aperture and the circle of least confusion

The lateral spherical-aberration blur radius grows as $b = c h^{3}$ for a constant $c$ . A lens gives a blur of $0.80 mm$ at full aperture $h_{0}$ . (a) What blur remains at $h_{0} /2$ ? (b) By what factor must you stop down to reach a $0.10 mm$ blur?

Recall Solution

(a) WHAT: use the cube law $b \propto h^{3}$ . WHY: the correction term is $θ^{3} \sim h^{3}$ . $b (\frac{h _{0}}{2}) = 0.80 \times (\frac{1}{2})^{3} = 0.80 \times \frac{1}{8} = 0.10 mm .$ (b) Set $b / b_{0} = (h / h_{0})^{3} = 0.10/0.80 = 1/8$ , so $h / h_{0} = (1/8)^{1/3} = 1/2$ . Stop down to half the radius (one aperture-radius halving = three f-stops in area terms).

L4.2 — Combine both aberrations

White light passes through the single crown lens of L2.1 ( $f_{D} = 200 mm$ , $ω = 0.01547$ ) which also has spherical aberration giving longitudinal SA $= 4.0 mm$ at full aperture. (a) Which is larger on-axis: the chromatic spread $f_{C} - f_{F}$ or the longitudinal SA? (b) If you stop the aperture to half radius, does the comparison change? Explain.

Recall Solution

(a) Chromatic $f_{C} - f_{F} = 3.09 mm$ (from L2.1); longitudinal SA $= 4.0 mm$ . At full aperture SA is larger ( $4.0 > 3.09$ ). (b) WHAT changes: longitudinal SA also scales as $h^{2}$ (it is $\propto h^{2}$ while lateral blur is $\propto h^{3}$ ). Halving $h$ gives $SA \to 4.0 \times (1/2)^{2} = 1.0 mm$ . Chromatic spread is unchanged by aperture (it is a focal-length shift, independent of $h$ ), so it stays $3.09 mm$ . Conclusion: after stopping down, chromatic now dominates ( $3.09 > 1.0$ ). This is exactly why fast lenses fight SA with shape and colour with a doublet — different levers for different heights.

Level 5 — Mastery

L5.1 — A parabola beats the sphere (numeric)

A concave mirror of radius $R = 400 mm$ has paraxial focal length $f = R /2 = 200 mm$ . For a marginal ray hitting at height $h = 60 mm$ , the spherical mirror's marginal focus falls short of the paraxial focus by approximately $Δ \approx \frac{h ^{2}}{2 R} \cdot \frac{1}{2} = \frac{h ^{2}}{4 R} .$ (a) Compute $Δ$ . (b) State why a paraboloid would give $Δ = 0$ for this on-axis parallel beam, and cite the one aberration it does not fix.

Recall Solution

(a) WHAT: plug numbers into the given longitudinal-SA estimate. WHY: it captures the leading $h^{2}$ falloff of the spherical mirror's focus. $Δ = \frac{h ^{2}}{4 R} = \frac{6 0 ^{2}}{4 \times 400} = \frac{3600}{1600} = 2.25 mm .$ The marginal rays focus 2.25 mm nearer than the paraxial focus. (b) A paraboloid is defined as the locus of equal optical path from an on-axis point at infinity to the focus, so all axial parallel rays arrive in phase at one point — zero SA, no $θ^{3}$ error. But it does not fix coma: off-axis parallel bundles still smear into comet-shaped blurs.

L5.2 — Full doublet design (open synthesis)

Design a cemented doublet of focal length $F = + 300 mm$ that is achromatic for the $C$ and $F$ lines. Given crown $ω_{1} = 0.020$ , flint $ω_{2} = 0.036$ . Find $P_{1}, P_{2}, f_{1}, f_{2}$ . Then compute the residual chromatic spread if the flint's $ω_{2}$ were mis-specified by $+ 5%$ (actually $0.0378$ ) — i.e. is the doublet still nearly achromatic?

Recall Solution

Step 1 — target power. $P = 1/ F = 1/0.300 = 3.333 D$ . Step 2 — solve the two conditions $P_{1} + P_{2} = 3.333$ , $ω_{1} P_{1} + ω_{2} P_{2} = 0$ : $P_{2} = - \frac{ω _{1}}{ω _{2}} P_{1} = - \frac{0.020}{0.036} P_{1} = - 0.5556 P_{1},$ $P_{1} (1 - 0.5556) = 3.333 \Rightarrow P_{1} = \frac{3.333}{0.4444} = + 7.500 D,$ $P_{2} = 3.333 - 7.500 = - 4.167 D .$ Focal lengths: $f_{1} = 1/7.500 = + 133.3 mm$ (crown), $f_{2} = 1/ (- 4.167) = - 240.0 mm$ (flint). Step 3 — sensitivity check. The residual colour power is $d P = ω_{1} P_{1} + ω_{2}^{'} P_{2} = 0.020 (7.500) + 0.0378 (- 4.167) = 0.1500 - 0.1575 = - 0.0075 D .$ Compare to a single crown lens of the same power, whose colour power would be $ω_{1} P = 0.020 \times 3.333 = 0.0667 D$ . The doublet's residual $∣ - 0.0075 ∣$ is about 11% of that — an order of magnitude smaller. Conclusion: even with a $5%$ glass error the doublet is still strongly (though not perfectly) achromatic. Real designs re-optimise or add a third element to kill this "secondary spectrum."

Recall One-line self-test (reveal)

Stopping down cuts spherical aberration but leaves chromatic untouched — true or false? ::: True; SA scales with aperture height $h$ , chromatic is a $λ$ -driven focal shift independent of $h$ .